Functional Equation: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $(x+y)(f(x)-f(y))=(x-y)f(x+y)$

$f(x)$ must be of the form $ax^2 + bx$.

Letting $x = 1$ and $y = 0$ in the equation, we find $f(0) = 0$.

Now define $g(x) = f(x)/x$ for $x \ne 0$. Taking $f(0) = 0$ for granted, the functional equation can be rewritten as $$(x-y)g(x + y) = xg(x) - yg(y), \qquad x, y, x+y \ne 0.$$

Substituting $1$ for $y$, we find $$(x-1)g(x+1) = xg(x) - g(1), \qquad x \ne -1, 0.$$

Now substituting $x + 1$ for $x$ and $-1$ for $y$, we find $$ \begin{align*} (x+2)g(x) &= (x+1)g(x+1) + g(-1) &\quad \text{($x \ne -1,0$)}\\ (x-1)(x+2)g(x) &= (x+1)(x-1)g(x+1) + g(-1)(x-1) \\ (x^2 +x - 2)g(x) &= (x+1)[xg(x) - g(1)] + g(-1)(x-1) \\ -2g(x) &=-g(1)(x+1) +g(-1)(x-1). \end{align*} $$ The last relation is in fact true for all $x \ne 0$, including $x = -1$.

This proves that the function $g(x)$ is linear, say $g(x) = ax + b$. Thus $f(x) = xg(x) = ax^2 + bx$. The relation $f(x) = ax^2 + bx$ is valid for all $x$, including $x = 0$.

Note that $f(x) = a x^2 + b x$ is a solution. I believe these are all the analytic solutions.

EDIT:

Yes, in fact they are all the differentiable solutions.

Taking $x=0$ we get $y f(0) = 0$, so $f(0) = 0$. Now suppose $f$ is differentiable. Taking the derivative of the equation with respect to $x$ and substituting $x=0$ we get $$ - 2 f(y) + y f'(0) + y f'(y) = 0 $$ Letting $f'(0) = b$, the solutions of the differential equation $-2 f(y) + b y + y f'(y) = 0$ are $f(y) = b y + a y^2$ where $a$ is arbitrary.

EDIT: Since $\dfrac{f(x) - f(y)}{x-y} = \dfrac{f(x+y)}{x+y}$, any solution that is continuous will be differentiable except possibly at $0$. I am not at all convinced that every solution must be differentiable at $0$, or indeed that every solution must be continuous.

Hint:

First take $y=-x/2$ then

$$f(x) = 3f(x/2) + f(-x/2)$$

Taking $x\to -x$ we get

$$f(-x) = 3f(-x/2) + f(x/2)$$

This motivates us to define $g(x) = f(x) + f(-x)$ which by adding the equations above is found to satisfy

$$g(x) = 4g(x/2)$$

which is much easier to work with. We can for example by a simple inductive argument show that $g(2^n x) = 4^n g(x)$ for all $x$ and $n\in\mathbb{Z}$. A simple consequence of this is that if we know $g$ on any interval on the form $(2^{-n-1},2^{-n}]$ with $n\in\mathbb{Z}$ then we know it for all other values.